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2 years ago

g What is the final velocity of a hoop that rolls without slipping down a 6.92 m high hill, starting from rest

Physics

1 answer:

liberstina [14]2 years ago

5 0

Answer:

The value is $v = 8.24 \ m/s$

Explanation:

From the question we are told that

The height of the hill is $h = 6.92 \ m$

Generally from the law of energy conservation we have that

$PE = KE + RKE$

Here PE is the potential energy of the hoop which is mathematically represented as

$PE = mgh$

KE is the kinetic energy of the hoop which is mathematically represented as

$KE = \frac{1}{2} * m * v^2$

And

RKE is the rotational kinetic energy which is mathematically represented as

$RKE = \frac{1}{2} * I * w^2$

Here I is the moment of inertia of the hoop which is mathematically represented as

$I = m * r^2$

and w is the angular velocity which is mathematically represented as

$w = \frac{v}{r}$

$RKE = \frac{1}{2} * m r^2 * \frac{v}{r} ^2$

=> $RKE = \frac{1}{2} * m r * v^2$

$mgh = \frac{1}{2} * m * v^2 + \frac{1}{2} * m r * v^2$

=> $v = \sqrt{gh}$

=> $v = \sqrt{9.8 * 6.92 }$

=> $v = 8.24 \ m/s$

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An electronic line judge camera captures the impact of a 57.0-g tennis ball traveling at 32.2 m/s with the side line of a tennis

mote1985 [20]

Answer:

average acceleration is 1.365 × $10^{4}$ m/s²

Explanation:

given data

initila speed u = -32.2 m/s

final speed v = 21.6 m/s

time taken t = 0.00394 s

solution

we get here average acceleration that will be express as

v = u + at ..........................1

put here value and we get

21.6 = -32.2 + a × 0.00394

solve it we get

a = 1.365 × $10^{4}$ m/s²

so average acceleration is 1.365 × $10^{4}$ m/s²

3 0

3 years ago

A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta

jeyben [28]

Answer:

$V_0$

Explanation:

Given that, the range covered by the sphere, $M$ , when released by the robot from the height, $H$ , with the horizontal speed $V_0$ is $D$ as shown in the figure.

The initial velocity in the vertical direction is $0$ .

Let $g$ be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. $V_0$ will remain constant throughout the projectile motion.

So, if the time of flight is $t$ , then

$D=V_0t\; \cdots (i)$

Now, from the equation of motion

$s=ut+\frac 1 2 at^2\;\cdots (ii)$

Where $s$ is the displacement is the direction of force, $u$ is the initial velocity, $a$ is the constant acceleration and $t$ is time.

Here, $s= -H, u=0,$ and $a=-g$ (negative sign is for taking the sigh convention positive in $+y$ direction as shown in the figure.)

So, from equation (ii),

$-H=0\times t +\frac 1 2 (-g)t^2$

$\Rightarrow H=\frac 1 2 gt^2$

$\Rightarrow t=\sqrt {\frac {2H}{g}}\;\cdots (iii)$

Similarly, for the launched height $2H$ , the new time of flight, $t'$ , is

$t'=\sqrt {\frac {4H}{g}}$

From equation (iii), we have

$\Rightarrow t'=\sqrt 2 t\;\cdots (iv)$

Now, the spheres may be launched at speed $V_0$ or $2V_0$ .

Let, the distance covered in the $x-$ direction be $D_1$ for $V_0$ and $D_2$ for $2V_0$ , we have

$D_1=V_0t'$

$D_1=V_0\times \sqrt 2 t$ [from equation (iv)]

$\Rightarrow D_1=\sqrt 2 D$ [from equation (i)]

$\Rightarrow D_1=1.41 D$ (approximately)

This is in the $3$ points range as given in the figure.

Similarly, $D_2=2V_0t'$

$D_2=2V_0\times \sqrt 2 t$ [from equation (iv)]

$\Rightarrow D_2=2\sqrt 2 D$ [from equation (i)]

$\Rightarrow D_2=2.82 D$ (approximately)

This is out of range, so there is no point for $2V_0$ .

Hence, students must choose the speed $V_0$ to launch the sphere to get the maximum number of points.

7 0

3 years ago

What is a

LiRa [457]

Answer:

I think there six points

4 0

2 years ago

On a hot summer day, 3.50 ✕ 106 J of heat transfer into a parked car takes place, increasing its temperature from 36.5°C to 44.4

anygoal [31]

Answer:

a) $\Delta s=443037.9747\ J.K^{-1}$

b) $\Delta s=31868131.8681\ J.K^{-1}$

Explanation:

Given:

amount of heat transfer occurred, $dQ=3.5\times 10^6\ J$

initial temperature of car, $T_i=36.5+273=309.5\ K$

final temperature of car, $T_f=44.4+273=317.4\ K$

We know that the change in entropy is given by:

$\Delta s=\frac{dQ}{T_f-T_i}$

$\Delta s=\frac{3.5\times 10^6}{(44.4-36.5)}$ (heat is transferred into the system of car)

$\Delta s=443037.9747\ J.K^{-1}$

amount of heat transfer form the system of house, $dQ=5.8\times 10^8\ J$

initial temperature of house, $T_i=23.5+273=296.5\ K$

final temperature of house, $T_f=5.3+273=278.3\ K$

$\Delta s=\frac{dQ}{T_f-T_i}$

$\Delta s=\frac{5.8\times 10^8}{278.3-296.5}$

$\Delta s=31868131.8681\ J.K^{-1}$

6 0

3 years ago

a parking lot is going to be 60m wide and 240m long which dimensions could be used please please help

Bezzdna [24]

Answer: The area of the parking lot is 14,400 meters squared.

Explanation:

We have the dimensions of the parking lot.

60m by 240m

The units used here are meters.

Now, if we want to know the area of the parking lot is equal to the product between the length and the width:

A = 60m*240m = 14,400 m^2

The area of the parking lot is 14,400 meters squared.

3 0

3 years ago

Read 2 more answers